Algebra Examples

$x^{2} = 8 x - 16$

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $8 x$ from both sides of the equation.

$x^{2} - 8 x = - 16$

Step 1.2

Add $16$ to both sides of the equation.

$x^{2} - 8 x + 16 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = - 8$ , and $c = 16$ into the quadratic formula and solve for $x$ .

$\frac{8 \pm \sqrt{{(- 8)}^{2} - 4 \cdot (1 \cdot 16)}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 8$ to the power of $2$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot 16$ .

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Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{8 \pm \sqrt{64 - 4 \cdot 16}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $16$ .

$x = \frac{8 \pm \sqrt{64 - 64}}{2 \cdot 1}$

Step 4.1.3

Subtract $64$ from $64$ .

$x = \frac{8 \pm \sqrt{0}}{2 \cdot 1}$

Step 4.1.4

Rewrite $0$ as $0^{2}$ .

$x = \frac{8 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Step 4.1.5

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{8 \pm 0}{2 \cdot 1}$

Step 4.1.6

$8$ plus or minus $0$ is $8$ .

$x = \frac{8}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{8}{2}$

Step 4.3

Divide $8$ by $2$ .

$x = 4$

Step 5

The final answer is the combination of both solutions.

$x = 4$ Double roots